If { }^{n-3} C_r + B cdot { }^{n-3} C_{r-1} + | vedclass

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(C) We use the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$. Given: ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_

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$(3, 3)$

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(C) We use the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$. Given: ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$. We know that ${ }^n C_r = { }^{n-1} C_r + { }^{n-1} C_{r-1} = ({ }^{n-2} C_r + { }^{n-2} C_{r-1}) + ({ }^{n-2} C_{r-1} + { }^{n-2} C_{r-2}) = { }^{n-2} C_r + 2 \cdot { }^{n-2} C_{r-1} + { }^{n-2} C_{r-2}$. Expanding further: ${ }^{n-2} C_r + 2({ }^{n-3} C_{r-1} + { }^{n-3} C_{r-2}) + ({ }^{n-3} C_{r-2} + { }^{n-3} C_{r-3}) = { }^{n-3} C_r + { }^{n-3} C_{r-1} + 2 \cdot { }^{n-3} C_{r-1} + 2 \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^{n-3} C_r + 3 \cdot { }^{n-3} C_{r-1} + 3 \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3}$. Comparing this with the given equation ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$,we get $B = 3$ and $B^{\prime} = 3$. Thus,$(B, B^{\prime}) = (3, 3)$.

If ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$ holds for all $n \geq r \geq 3$,then $(B, B^{\prime}) = $.

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